geometry-and-trigonometry

[upper karma]

first method

the law of sines/law of cosines is suited for ALL triangles, not just right triangles

it's pretty useful to know

[wooden current]

area segment = area of sector - area of triangle = $\frac{\pi*\theta}{360}*R^2 - \frac{R^2*sin(\theta)}{2}$

[somber coyote]

DanCastan:

[wooden current]

Factor out $R^2$ and you just plug in the values

[somber coyote]

DanCastan:

[wooden current]

@dim wren Do you still need help?

The measure of BAC is half the measure of BOC

[acoustic jungle]

yes

[upper karma]

@upper karma the dark colored part from here? https://discordapp.com/channels/268882317391429632/326138757474680852/710570066198134864

also, is this a test?

@upper karma is this an exam?

[lunar sand]

i need some guidance

[acoustic jungle]

find circumcenter

[lunar sand]

Is there a formula for that? I haven't learnt what circumcenter is.

[vague berry]

might need some help

stand by

nvm im good

[kind temple]

@lunar sand That question is wrong I believe

https://discordapp.com/channels/268882317391429632/326138757474680852/710553916500213782

This was the circle made by @upper karma

Plotted

And it lies in the 1st quadrant so both of the co-ordinates of the centre have to be positive but they arent

I also tried the question in a different way

By assuming every option as the centre and calculating the radius but I never got the same radius for every distance in any option

Here https://discordapp.com/channels/268882317391429632/326138757474680852/710559211796299887

[lunar sand]

well, I'll just get a few more opinions on the question before deciding that it's an impossible question.

[kind temple]

It's not impossible

The options are wrong

[lunar sand]

ah, yeah. I didn't mean it that way.

[kind temple]

Do you want to solve it regardless the options?

If you want to do it that way then you can take the coordinates of the centre as (x, y)

[upper karma]

@upper karma did you also get the same result?

[kind temple]

And then get 3 equations of centre by distance formula

Equate them and get it

[upper karma]

@upper karma rounded to the nearest tenth would be 13.9

[shell merlin]

could someone help i cant fine how to do this i dont need the answer just a formula i can use or a link to a khan academy vid on this

[upper karma]

@upper karma

,w 2x+11 = 4x-3

[somber coyote]

Display results online and refine query

Requested by DSpider#6917

[upper karma]

^this was question 24

[upper karma]

@upper karma https://www.youtube.com/results?search_query=circle+theorems

Let’s

[humble pebble]

[upper karma]

@upper karma Yes son

Circle theorems deal with stuff within a circle

[wooden current]

@shell merlin Use Intersecting Chord Theorem

[upper karma]

@shell merlin how long is the radius?

and would you say it's the same as this?

@shell merlin and can you figure out half of AB?

hint: ||Pythagora||

@upper karma did you go through those videos?

[wooden current]

What about Intersecting Chords Theorem 😭

[upper karma]

🙂

[wooden current]

😳

[shell merlin]

thx im good @upper karma

[wooden current]

epic

[upper karma]

@upper karma which one?

which one from questions 25-32?

@upper karma so you need help with all of them except 28 and 31

@upper karma what can you tell me about angle BAX?

it's an inscribed angle

how about now?

what can you tell me about that angle now?

what's the measure of angle BAX?

yes

how long is arc EX?

remember, the whole circle is 360 degrees

yes

and are there any angles we could use with this arc?

180 + 100 + 80 = 360

sure

yep

if that's the new diameter, what is the measurement of angle ABC?

so that takes care of questions 25 and 32

https://youtu.be/GppOSNTi5OA?t=718

Circle theorems are pretty awesome

Ngl

Oh

Learn to love them

did you see the above video?

good

https://youtu.be/GppOSNTi5OA?t=744

click the video

it has a timestamp

yes

the guy even explains what an inscribed angle is

(all the vertices of the triangle are on the circumference of the circle)

half, yes

so that would be question 26

[agile tree]

Would anyone be willing to help me?

[upper karma]

@agile tree I'll be with you in a sec

[agile tree]

Thanks spider

[upper karma]

CBD is the same as CED, yes

CBD is the same as CED, yes

[agile tree]

Im still here waiting spider just lmk when u can help

[upper karma]

@agile tree I'll ping you

@agile tree i can

Lets go to another channel tho

[agile tree]

Dm me @upper karma

[upper karma]

public is better

No, on #prealg-and-algebra

so other people can pitch in and correct

Yeah

I said it bc of bots lol

But ok

[agile tree]

ok thanks

well go to pre alg

[thorn holly]

anyone know about coterminal angles

[upper karma]

@upper karma you got them all

good job

I was about to ask you how long is arc AD

but you got it right

well, what is the measure of this angle?

?

ok

yes

it's an inscribed angle

so you know all angles from a triangle add up to 180 degrees?

and if that angle is 90... and the entire equation needs to equal 180...

they other two add up to 90

they're not necessarily both 45 degrees

you didn't watch the above video 😦

you would add those expressions + 90 = 180
yes

(1/2)x + (1/3)x + 5 + 90 = 180

solve for x

,w (1/2)x + (1/3)x + 5 + 90 = 180

[somber coyote]

Display results online and refine query

Requested by DSpider#6917

[upper karma]

🙂

no problem

[thorn holly]

how do you find the quadrant of a negative angle?

[upper karma]

@thorn holly google "unit circle quadrants"

[versed river]

Thales theorem and angle sum of a triangle should help

[upper karma]

@upper karma https://youtu.be/XWqP9T7-HXM?t=907

[upper karma]

@upper karma did you get it?

what answer did you get?

[prime barn]

can anyone help me with this problem about systems of equations?

dm if you can

[upper karma]

sure, no problem

I'm off to bed

@prime barn post your question if it's related to trig

maybe someone can help

[prime barn]

well its a weirder question where you need custom equations

[upper karma]

if it's not related to trigonometry/geometry, maybe post in one of the #help-2 [greek_letter] channels

[prime barn]

here ill just put it and see

[shell raft]

In order to find the height of a tree, would it be good enough to get a protractor to measure from the ground to the top of the tree? Also, I will measure the distance from the base of the tree to where i'm standing. I can't afford expensive gizmos. If someone thinks there is a better way, please suggest ideas. Thanks.

[upper karma]

@shell raft you don't necessarily need to be on the ground... you can put it on top of a vertical stick that has a fixed height (e.g. 1 m)

once you have the angle you need to know the distance from you to the tree

and then just take an angle measurement from you to the top of the tree

then, because you have 2 angles and a distance (one angle you measured, the other is 90 degrees), you can easily get the height of the tree

[shell raft]

thank you dspider.

[upper karma]

@shell raft

g'night

[shell raft]

thank you.

[storm silo]

is this right?

Anyone?

<@&286206848099549185>

[shell raft]

the answer is correct.

[acoustic jungle]

is the bottom not included?

Also not to be that guy but if you sub pi as 3.14 the answer is 18.84 not 18.85 lol

[shell raft]

sorry, that is my fault. i didn't use 3.14 and i rounded at the end so i assumed roku did that as well.

[kind glen]

Could someone possibly assist me with this?

Find a Ferriswheel. State name and location. Find length of time for a rotation and the diameter. If it does not mention where people board assume it is at ground level. Create a table giving height based on time have the correct units. Find midline amplitude and period and from this info get formula for the function that gives you height.

I'm using the High Roller which is 550ft tall with a diameter of 520ft and has a total rotation time of 30 minutes

[acoustic jungle]

@kind glen midline is 30+radius, amp = radius, 2pi/30 = k, or the stretch/compression. cos starts at the top so you have to do k(x-half the period) or just put a negative infront of a, where a is the amplitude in acos(k(x-d))+c

[upper karma]

how is this incorrect

You multiplied wrong.

Answer is an integer.

[upper karma]

https://gyazo.com/ef3bf75ef8371e5cbfaf4204dbf50479

how do i go about solving this

[dark sparrow]

what's giving you trouble here

[upper karma]

v = bh

i don't know the steps to take

[modern mountain]

find the area of a cross section

[upper karma]

find the area of the base

the base and height are already given

the lengths

so

[azure reef]

look

you have height of triangle and its base

[upper karma]

yes

9 and 6

[azure reef]

do you know the formula for triangle's area?

[upper karma]

A=hbb
2

yes

A = 21

correct?

[azure reef]

so it would 18 yd squared

[upper karma]

.

[azure reef]

oh

[upper karma]

21 yd squared?

[azure reef]

i am idiot sorry

9 yd is height

and 6 is base

[upper karma]

27 yd ^2

[modern mountain]

27?

[azure reef]

then area is
yes 27

[upper karma]

okay now what

do i do with the area

[azure reef]

then you multipliy this are by the width of base of prism

which is 8 yd

and get 27*8=216

[upper karma]

right

[azure reef]

so the volume is 216 yd^3

[upper karma]

216 yd ^3

why cubed?

[azure reef]

because

unit of length times unit of height times unit of width us yd^3

yd times yd times yd is yd^3

[upper karma]

ah okay i remember that

okay

https://gyazo.com/0072b01b087fe078ca53486cfe2c7305

im going to do this one

lets see if i can do it right

175 yd^3

[azure reef]

yep

[upper karma]

finally

someone helps me and simplifies it

thank you so much

ill finish this up and

i have more mabye

if you have time

don't feel obligated

thank u

[azure reef]

you are free to write in private message if you need or to ask someone another, i think someone would anyway help you

no worries

[upper karma]

https://gyazo.com/0baa7b8350de1af85f89c3dd126596eb

so far i have 18 * 28 = 504

what are the next steps/formula

[versed river]

formula for a circle area is A=pi*r^2. what you ahve there is a semicircle, which is half a circle

so it would be half of that formula

[upper karma]

hmm

how do you have the radius though

[shell raft]

the diagram didn't give you the specific radius.
it gave you something else to figure out the radius. Do you notice what that might be?

[upper karma]

It gave me the rectangle.

@shell raft

[versed river]

yes. can you use one of the values from that rectangle to figure out the radius?

[upper karma]

I'm not sure

@versed river

[versed river]

notice these two sides are the same length

since its a rectangle

i.e theyre both 18 ft

does that make sense to you?

[upper karma]

ahh

yes

[versed river]

do you see how to get the radius from there?

[upper karma]

i cant seem to find

my formulas

mind helping me with this one

then ill try working on my own

so whats the forrmula

[versed river]

formulas you need here are area of rectangle=lw where l is length and w width, and area of a semicircle=(pi*r^2)/2

[upper karma]

ah

okay

so

3.14 ( )^2/2

[versed river]

yes

() being the radius

[upper karma]

how about now? do you see the radius now?

9?

of course

ahhhhh

right so,

3.14 (9)^2/2

28.26

is what i have here

[versed river]

• the area of the rectangle

[upper karma]

ah

yeah

i gotta add that

[versed river]

,w (3.14* ((9))^2)/2

[upper karma]

74.26

is the answer

74.26 ft^3

[somber coyote]

Display results online and refine query

Requested by Sneaky#0928

[shell raft]

wait. whay ft^3?

are you sure about the units?

[upper karma]

Area?

[versed river]

ft^3 would be for volume

[upper karma]

right

ok

squared

your right

you're*
(you + are = you're)

oh my answer

was way off

[versed river]

and 74.26 is wrong

[upper karma]

oops

well, no

now

it's not letting go back

its on this practice website

https://gyazo.com/75ebb99cba3d7ce5d38a96926b436f2e

this shit is so hard for me

for whatever reason

[versed river]

can you find the area of the semicircle?

not worrying about the triangle yet?

[upper karma]

56.549

is the area

[versed river]

of the semicircle?

or the thing with the triangle cut out

[upper karma]

semi

circle

[versed river]

that doesnt sound right

[upper karma]

would be

[shell raft]

uhh danny. your answer is close but it is not right. did you use 3.14 for pi?

[upper karma]

yes

i believe so

[versed river]

,calc (1/2)3.146^2

[somber coyote]

Result:

56.52

[upper karma]

right

[versed river]

how did you round it?

[upper karma]

say again?

its 3 am for me

im about to give up for the night

[versed river]

your answer was 56.549. did you do (1/2)*3.14 *6^2 to get that?

[upper karma]

yes

i believe so

[versed river]

did you use a calculator? it should be 56.52

[upper karma]

i messed up the pi

[versed river]

are you sure you used 3.14 and not pi?

ah

[upper karma]

somehow

[versed river]

thats probably it then

[upper karma]

i used 3.13

3.14

not pi

ok

anyways

[versed river]

so now we wanna get the area of the triangle right?

[upper karma]

yes

[versed river]

do you know the formula for that?

[upper karma]

@upper karma

A=hbb/2

[versed river]

what are the variables here

not sure what bb is

[upper karma]

base and height

[versed river]

ok so base * height/2 yeah

[upper karma]

yeah

[versed river]

can you recognise what those are?

[upper karma]

yes

[versed river]

so they are ?

[upper karma]

base would be 18.84

[versed river]

how did you get that?

[upper karma]

wait

i didnt divide by 2

it would be

OH

WAIT

113.04‬

then divided by 2?

[versed river]

uh

i dont think so

[upper karma]

56.52‬

no.

[versed river]

how did you get that...

[upper karma]

hmm.

i took the area we had

times it by

[versed river]

h is 6 feet, thats given

[upper karma]

okay

[versed river]

b is the same as the diameter of the circle, right?

[upper karma]

yes

[versed river]

if the radius is 6, the diameter is _________?

[upper karma]

119

Sorry

[versed river]

?

[upper karma]

What

12

[versed river]

...

thats correct but let me explain why first....

remember the diameter is twice the radius

[upper karma]

Because a radius is from center to circumference

[versed river]

and we know the radius is 6 ft

so the diameter is?

[upper karma]

3

And a diameter connects 2 end points of the circumference

i mean

And passes through the middle

[versed river]

diameter is twice the radius

[upper karma]

Idk

Yes

12

yeah

[versed river]

yeah

so now we can just plug that into our formula for area of a triangle

[upper karma]

okay

[versed river]

(base times height)/2

base is 12, height is 6

[upper karma]

72

[versed river]

/2

[upper karma]

/2 = 36

[versed river]

yup

[upper karma]

okay

[versed river]

now whats the shaded area, in terms of the area of the semicircle and the area of the triangle?

https://images-ext-2.discordapp.net/external/roEhVr8O_CB3vTthZUYxQVcJLA8yYkEEN46akC9ieZ0/https/i.gyazo.com/thumb/1200/75ebb99cba3d7ce5d38a96926b436f2e-png.jpg

[upper karma]

its a circle

?

[versed river]

mm the shaded area is the area of the semicircle minus the area of the triangle right?

[upper karma]

yes

essentially

[versed river]

the area of the semicircle is 56.52

the area of the triangle is 36

[upper karma]

correct

[versed river]

and the shaded area is the semicircles area minus the triangles area

[upper karma]

mm

[versed river]

so whats that?

[upper karma]

16.52

if you subtract that

shaded area

area of semi from area of triangle

[versed river]

,calc 56.52-36

[somber coyote]

Result:

20.52

[upper karma]

ok

[versed river]

you lost a 4 somewhere

[upper karma]

im sorry..

its 3:15 am

im having troubles staying awake

but i need to finish this

so we have 20.52

after subtracting

[versed river]

yes. thats the area of the shaded region, agree?

in ft^2

[upper karma]

yes

ft squared

[versed river]

so thats your answer?

[upper karma]

20.52 ft ^2 is my final answer

[versed river]

sounds good

[upper karma]

thanks

alot

❤️

[atomic delta]

Are there any elliptic theorems in angles just like with circle theorems but with ellipses?

[upper karma]

well, an ellipse is just a circle with one of its axis slightly dilated or stretched

I think the same theorems for circles should apply, as long as you account for that stretch/squish ratio

[dark sparrow]

uh

you know that stretches fuck angles up royally, right

[azure reef]

well, an ellipse is just a circle with one of its axis slightly dilated or stretched
more correctly is to say that a circle is an ellipse with equal major and minor axes

[pallid cloud]

@azure reef and an ellipse is a circle with one of its axis slightly dilated or stretched is correct too

Consider a transformation called orthogonal affinity

Let its direction be parallel to one of the axis of the ellipse

And its ratio be the ratio of the lengths of the axis

Then it transforms the ellipse to a circle

Both are correct

You van transform a circle toban ellipse in this transformation

And an ellipse to a circle too

[azure reef]

sure, but i mean that from the point of view of analytic geometry ellipse is more general case

[pallid cloud]

You can see a circle as an ellipse but with focii the same point, the center of the circle

Yes sure

Ellipse: x^2/a^2+y^2/b^2=1
Circle x^2/a^2+y^2/a^2=1
The ellipse is more general yes I see what you mean

[azure reef]

anyway, it is just terminology, one can use this terms interchangeably except for specific problems

[pallid cloud]

Yes

We can go from circle and the generalize to ellipse, analytically or geometrically

Or we can go from the general case, ellipse, to the particular case, circle

[mellow sluice]

what does it mean to find the equation of a line such that the composition Rx-axis ∘ Rl = R0,90?

I understand Rx-axis would be reflecting the line upon the x-axis

what does Rl and R0,90 mean?

[pastel anchor]

if im trying to find the area of parallelogram and got the height, but there are two different bases.. which one do I use?

[pallid cloud]

@pastel anchor aren't the bases equal?

The opposite of course

[pastel anchor]

let me send a pic

[pallid cloud]

Ok

[pastel anchor]

the question asked for the area of the shaded polygon

[pallid cloud]

This is not a parallelogram

The shaded region is limited by a quadrilateral that is not a parallelogram

[pastel anchor]

so what is the formula to shade it?

the textbook lesson was saying the formula was a=bh

[pallid cloud]

For a parallelogram yes

[pastel anchor]

well im suppose to find the area of the shaded polygon, so do i d 16*10?

[pallid cloud]

No

[pastel anchor]

so how can i do it

[pallid cloud]

You can see the shaded region as a parallelogram and a triangle next to each other

So the area of the shaded region is the sum of the area of that parallelogram and the area of that triangle

[pastel anchor]

the triangle on the right?

[pallid cloud]

You know that for a prallelogram the area is given by A=base×height and for a triangle the area is A=base×height/2

Apply this to this figure

No not the triangle on the right for sure, because it's not shaded

The triangle on the left

[pastel anchor]

how can i find the height for the triangle on the left?

[pallid cloud]

It is the height given !

For the parallelogram and the triangle

h=10m for both the parallelogram and the triangle

[pastel anchor]

ok so the area of the parallelogram is 110

[pallid cloud]

Just pick the bases for each to match this height

True

[pastel anchor]

and the area of the triangle is 25?

so do we add it to find the area of the shaded?

[pallid cloud]

True

[pastel anchor]

alright

i don't get this one

because it has no height

[pallid cloud]

Do you know Pythagoras' theorem?

[pastel anchor]

yes

[pallid cloud]

Use it here

[pastel anchor]

ohh

[pallid cloud]

To find the height that is common for eachvof the two triangles and the parallelogram

[pastel anchor]

so the height is 4

16

actually

[pallid cloud]

You should get 24 I guess

[pastel anchor]

wait i used the 10+x=26

[pallid cloud]

This is not Pythagoras' theorem

[pastel anchor]

but arent they already squared?

[pallid cloud]

10^2+x^2=26^2

No

A length is not squared

When it's squared it is an area

[pastel anchor]

so i combined all the areas of the shapes

and i got 1200m

[pallid cloud]

m^2

And I will check the number

10×24/2=120

20×24/2=240

40×24=960

[pastel anchor]

i think i made a mistake in the 2nd triangkle

[tacit karma]

its a trapezoid.-.

[upper karma]

yoh

anyone here able to help me with a problem i got

https://gyazo.com/e40d656117a80e48451d12fd38c1ba18

i got no clue how to work the problem

[acoustic jungle]

you can find "a" by pythagoras

[upper karma]

@pastel anchor

how about now?

[acoustic jungle]

then use pi * a^2 to find the area

[upper karma]

"Given that the tent is a spherical radius of..."

I'm sorry, what?

the tent is a spherical radius?

[upper karma]

@acoustic jungle what

explain

[acoustic jungle]

Find a using r^2-(r-h)^2=a^2

[silent plank]

@upper karma AO is also a radius of the circle
the length of CO can be obtained from r - h
and then apply pythag to get a

[dense sky]

can someone help me

[upper karma]

r^2-(r-h)^2=a^2

whoops

dude why am i still lost...

[tacit karma]

@dense sky what do you know about tangents of a circle

[upper karma]

co, ob?

[dense sky]

@tacit karma I know that the sides are equal

like 20 goes on the other side

[tacit karma]

so what is the length of CD

[dense sky]

20

[silent plank]

capitals, the notation implies the length of those line segments

[tacit karma]

and the length of NW

[dense sky]

20?

[tacit karma]

no...

[silent plank]

can you move to #help-1 where you also posted the same question

this is in use

[upper karma]

to be fair

i barely understand this image

[tacit karma]

so CZ is equal to ZN right

[upper karma]

im so gnna fail this

fucking 2 hours on this one question

[dense sky]

yeah

[tacit karma]

so ZN= to what?

[dense sky]

20? ;-;

[tacit karma]

yes

and NW is what then

[silent plank]

the image is basically a cross section of the middle of the dome

[dense sky]

i dunno what NW is..

[upper karma]

dude

omg

@upper karma fail what?

is this a test?

if i miss 1 question on this i fail the semester

[tacit karma]

camden go to #help-1

[upper karma]

no its not a test

we have these learning progress checks every week, i got mono and recently recovered so ive had to do 1 months worth of work this week

and im not able to get help from teachers for atleast 2 days and by then it will be over

[silent plank]

and the cross section of that dome would be part of a bigger circle

since they nicely provided the diagram for you, you can ignore most of the text and just use that

[upper karma]

i legit never got a video tutorial on these

ive had to search everything up and some shit idek what to search up

so is the r the 14.1 ft

[silent plank]

radius of a circle
properties of addition
pythagoras

no

[upper karma]

wtf is it then

[silent plank]

wait

sry yes.

[upper karma]

...

[silent plank]

got my things mixed up.

[real lily]

Do i do 8x+4=9x-5 and go from there?

[upper karma]

then how am i suppose to find co

[real lily]

not sure why but this is confusing me

[silent plank]

in use

move to an open chanel

do you see that AO is also a radius of the circle?

[upper karma]

yeah

[silent plank]

and that:
OC + CA = AO = r

in use
move to an open channel

[upper karma]

@upper karma

Im sorry, which channel?

[real lily]

^

[upper karma]

any of the #help-2 [greek letter]

wait

so its =\sqrt(14.1^(2)-6.1^(2))

that didnt want to show up pretty

[real lily]

@upper karma so just put my problem in any of those channels?

[upper karma]

yes

if it's free unocuppied

@upper karma so what's the length of OC?

the same as ac?

look at the image I posted above

wait

the red question mark

yeah

do you see it?

[silent plank]

OC + CA = AO = r

[upper karma]

oh

alr

8.8

...

read that num wrong

ITS Hard to read

it is, isn't it?

..

8

look at the problem's requirements

yes

ok

so you've got 14.1 - 6.1 = 8

good

now do you see where Pythagoras' theorem could be applied?

yh

a = ?

11.61077086

,w a^2 + 8^2 = 14.1^2

this shit is hard without knowing how to do it, any of the terms, and no fucking teacher

[somber coyote]

Display results online and refine query

Requested by DSpider#6917

[upper karma]

ok

ah shit

the next one is on the same tent

fuck this tent

"To the nearest foot how tall could the tent be if a group of campers needed 127.7π square feet of ground space in a tent with a spherical radius of 16.9-feet?"

if you round it to the nearest hundred, it's 11.61 ft

yeah i put that

sorry, but could you help me on this one

ok, so the area of this new hemisphere (?) is 127.7 pi ft^2, with a radius of 16.9

why do you think it's the same tent?

no, same image type shit

post it

let's see

https://gyazo.com/4461942fb46a292143f9742858d4122d

ok, so square feet is the area

area of a circle

do you know the formula?

(google it)

of the area of a circle?

yes

yeah

wait

or maybe you don't need it

do i need to find out the radius through that

just take the square root of 127.7

and i would get the radius

but they already give you the radius?

or do they give you the radius of the sphere?

radius of the sphere

spherical radius

yeah

it's the sphere

carry on

you're doing great

so you have the area for the ground floor (which is a circle)

did you google the formula?

yeah

oh

did i google it?

no

what's the formula for the area of a circle?

i had to use that formula in a diff problem earlier

pie r squared

pi

good

but its asking about height?/

so you have pi r^2 = 127.7pi

solve for r

we'll get to the height in a moment

11.30044246

right?

i mean i understand how to get the radius i just dont know what that has to do with the question

yep

ok, so the radius of the ground floor (which is a circle), is 11.3

using Pythagoras' theorem, can you solve the red question mark?

um

no

(which kinda looks more like a fat exclamation mark, lol)

you would need the height in order to do that

how about now?

oh

im dumb

already got 2

🙂

it's 2?

you only need 2 sides for it...

thats what im talking about

where did you get "2" from?

the a and r

you need 2 sides for Pythagorean theorem?????

uh, yes?

of course

bruh

what do you mean ofc, you asked where i got 2 from

i was talking about there already being 2 sides

...

,w x^2 + 11.3^2 = 16.9^2

[somber coyote]

Display results online and refine query

Requested by DSpider#6917

[upper karma]

yeah

got that

actually

not rlly

i got 12.56622458

I thought you meant you calculated it... ok, whatever

so OC = 12.6

yeah

[tacit karma]

... I tried to solve this problem and it had 2 solutions

[upper karma]

okay whats next???

h = ?

4.33377542

rounded to the nearest tenth?

its to the nearest foot

rounded to the nearest foot, actually

yeah

yh

4

4, yes

there you go

congratulations

[silent plank]

not nearest

[upper karma]

ree thats so dumb

[silent plank]

you'd need to round up

[upper karma]

no

[silent plank]

otherwise there wont be enough area

[upper karma]

4.3

oh

hm

i doubt it

i dont think the question cares about life

[tacit karma]

the answer could be 29, but the tent would be weird

[upper karma]

Can i ask my question now please?

yeah

@silent plank 4.3 is nearest?

🤔

[silent plank]

5 would be the nearest foot

[upper karma]

its not about the real world though

[silent plank]

well the minimum integer foot value

it is though

[upper karma]

I would still put it down as 4.3

im doing 4

16.9 - 12.6 = 4.3

it will tell me if its wrong and if it is i can just do the same thing with different numbers

4 does make more sense (to me at least)

oh, cool

there you go then

same since its not a real world application

[tacit karma]

isn't 29 technically still correct?

[upper karma]

let us know which value it takes

well i still got 1 more

ima try to work on it myself rn

ok

good

Is anyone able to help me with my question in question room?

[tacit karma]

which question room @upper karma

[upper karma]

umm, it is the 4th one

LOL

it took it

[tacit karma]

lol

[upper karma]

it took the whole number?

it took 4

[tacit karma]

Anyone wanna try a question i made myself?

[upper karma]

no im still braindead

closing in on 3 days no sleep

go sleep

no

i cant

i need to finish my classes

[tacit karma]

im barely doing classwork lol since they're not graded .-.

[upper karma]

they are for me

i need atleast 3/4 to pass the semester on everything

i had mono for a month so when i got back to healthy, i ended up having 150 missing assignments

started on Tuesday and I'm almost done with them all

[tacit karma]

oh .-.

your teacher isn't excusing you?

[upper karma]

no

[tacit karma]

oof

[upper karma]

and i didnt get any note videos

so ive had to do this all without learning the vocab or material

[tacit karma]

that really unfair

[upper karma]

yeah

lifes unfair

and for some reason it always picks me

[tacit karma]

did ur parents email the teacher?

[upper karma]

ive emailed them all myself with a doctors note

apparently i couldn't get absent time off since its online school and not going to the actual school.

[tacit karma]

that really got to suck

[upper karma]

eh

i have a friend that helped do some of my essays

she fucking amazing at them. shes like at stanford right now studying ChE

[tacit karma]

thats nice

are you in high school

[upper karma]

last thing she sent me

" working in a bioengineering research lab with a master's student. she had just begun her thesis so i was able to work with her on her research and later become published. we worked with mesenchymal stem cells in hydrogel scaffolds and their abilities to speed up wound healing. the experiments were very time-sensitive so i found myself in the lab, often by myself, at weird hours in the evening when everyone else had left. that was by far my favorite part of senior year."

yeah im in highschool rn

[tacit karma]

oof

[upper karma]

eh

[tacit karma]

i feel very unproductive

[upper karma]

im graduating early

lol its alr

sometimes you need to be unproductive to give you that motivation to be productive.

ree

ive been trying out this thing with ego depletion

like I've been trying to make myself believe I don't need breaks and so far I'm 18 hours in without one.

[upper karma]

@upper karma what have you tried

[full ruin]

can i get a formula or something

i forgot the steps

[acoustic jungle]

perpendicular slope of a/b = -b/a

[full ruin]

i-

english please

ok someone just give the answer 😌

LMAO

[upper karma]

ok someone just give the answer 😌
@full ruin hell no, we dont do that here

[full ruin]

LMAOO

ok teach me them

then i’ll learn

i’m a fast learner

is it the third one @upper karma

i tried plugging in the points they gave me cause i had nothing else to try

[blazing meadow]

you can use POINT SLOPE FORM along with the perpendicular slope

if you recall that slope is:

you can rewrite it:

[full ruin]

ooo yes i understand this one

[blazing meadow]

you can replace the m with your slope and the x1,y1 with your given points

then turn it into a nice looking [y=mx + b] equation

[full ruin]

god bless 😌

✨✨

[blazing meadow]

the other option would have been looking at those 4 answers, determining which of them used the correct perpendicular slope. (the top two choices)
and then plugging the point in and seeing if it satisfies the equation (is the point in the line? )

[lament harness]

https://cdn.discordapp.com/attachments/187020781094567936/710834884507336774/20200515_204325.jpg

[weak shoal]

@lament harness Were you planning on asking a question or just posting the picture as decoration?

[lament harness]

why would I post a picture for decoration??

oh yea I forgot the question

I need help on c)

[weak shoal]

What help do you need? What have you tried? Were you able to do parts a) and b) successfully?

[lament harness]

yes

but im not sure how a) and b) can help with c)

[weak shoal]

Well, I assume that you have:

$\sin^2(C) = \sin^2(A)+\sin^2(B)+2\sin(A)\sin(B)$

$\cos^2(C) = \cos^2(A) + \cos^2(B) + 2\cos(A)\cos(B)$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Yes?

[lament harness]

yes

[weak shoal]

So:

$\sin^2(C) + \cos^2(C) = \sin^2(A) + \cos^2(A) + \sin^2(B) + \cos^2(B) + 2\sin(A)\sin(B) + 2\cos(A)\cos(B)$

Yes?

[somber coyote]

Abhijeet Vats:

[weak shoal]

I just added the both of them

[lament harness]

alright

[weak shoal]

So:

$1 = 1 + 1 +2[\cos(A)\cos(B) + \sin(A)\sin(B)]$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Can you use that to determine A-B

[lament harness]

ahhh

I see

thanks for it

[weak shoal]

You're welcome.

[split verge]

Do trig identities still apply to inverse trig functions?

[dark sparrow]

no

[split verge]

ok, thank you

[upper karma]

well, try applying one

sin(theta) + cos(theta) = 1
...for which their inverse would be:
sin^-1(theta) + cos^-1(theta) = 1

regardless of angle, they won't ever equal 1

[acoustic jungle]

sin(theta) + cos(theta) = 1
wtf

[high zephyr]

lmao

sin²(theta) + cos²(theta) = 1

[ripe dove]

wtf
@acoustic jungle well, it’s a conditionally true equation, just not an identity

Also, while simply replacing the trig function in the identity with its inverse function is likely what saif was meaning, that’s not what the inverse of a trig identity would be

[candid terrace]

Hey

I need help with an assignment due at 12:00

[upper karma]

Nice.

@candid terrace Ya I’ll help

Don’t hesitate to ask

Send it here

[candid terrace]

Ok

[upper karma]

Oh yes yes this is awesome

@candid terrace What have you tried so far?

[upper karma]

@candid terrace Hint, angle TAB and angle TCB are right angles. You have 2 sides of the triangles, what's stopping you from getting the 3rd?

Yea they are right angles because the angle between a tangent and radius is 90

That’s a theorem

[candid terrace]

Thanks guys

Wait

[upper karma]

Ye?

[candid terrace]

I’m still having trouble understanding what I should be doing

I’m finding the perimeter

[upper karma]

You know the radius right?

So you know that the two sides inside the circle are both 8.

[candid terrace]

8

[upper karma]

What you need to do is to find line CB and AB.

[candid terrace]

OOOOHHH

I get it

[upper karma]

👍

Congrats :3

[candid terrace]

I feel so stupid, I was staring right at it

[upper karma]

If you're stuck, just read the question then ask yourself: "What does this information tell me?"

They don't tell you that certain lines are tangent for the hell of it.

[candid terrace]

But hold on

[upper karma]

Ye?

[candid terrace]

What is the length of AB

Double the radius?

[upper karma]

You need to find it yourself.

[candid terrace]

Imma look up circle theorems

[upper karma]

Nonono.

Stay here.

You have two right triangles that are shaped like this.

Use the Pythagorean theorem to find the side.

@candid terrace

17 is that center line.

Pretty awesome

Lol yes.

[candid terrace]

Ok

It’s 15

🥳

Ok

All seriousness

This assignment is due at 12:00

I have 30 minutes left

But

It’s says the perimeter of ABCT

@upper karma

So is the answer 46?

[upper karma]

This is pretty easy

Wait

[candid terrace]

Since I’m not counting the hypotenuse

[upper karma]

Ya u got it

Yep

[candid terrace]

See

[upper karma]

Ya ik

[candid terrace]

ABCT

Ok

Please stay and help with the rest

This assignment is due in half an hour

[upper karma]

Ya I will

I got u fam

Send me some more

I can take it

We are brave men

[candid terrace]

Thanks bro

I’m lost @upper karma

[upper karma]

@candid terrace This problem is the same as the last one.

Except you have to draw some extra lines.

Draw a line from T to J.

And then you have an exact replica of the last problem.

Two right triangles next to each other and you have a shit ton of sides to go off of.

[candid terrace]

Oh

I see it know

[upper karma]

Mhm.

[candid terrace]

Jp 8

[upper karma]

Yes

[candid terrace]

JT 12?

[upper karma]

No

[candid terrace]

Then what is it?😥

[upper karma]

See how it forms a right triangle

JMT

Just find the hypotenuse

[candid terrace]

Ik

But

[upper karma]

You probably did your arithmetic wrong

[candid terrace]

I can’t find the length of the hypothenuse

[upper karma]

Yes U can

Mt Is 5

JM is half of JK

12

[candid terrace]

It’s 13

[upper karma]

Yea

[candid terrace]

But that’s the answer

[upper karma]

?

[candid terrace]

My bad

But that’s not the answer

It’s asking for the length of TP

[upper karma]

Yep

[candid terrace]

How do I find that

[upper karma]

By drawing line JT, 2 right triangles are formed